HDU 4925 Apple Tree

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Problem DescriptionI’ve bought an orchard and decide to plant some apple trees on it. The orchard seems like an N * M two-dimensional map. In each grid, I can either plant an apple tree to get one apple or fertilize the soil to speed up its neighbors’ production. When a grid is fertilized, the grid itself doesn’t produce apples but the number of apples of its four neighbor trees will double (if it exists). For example, an apple tree locates on (x, y), and (x - 1, y), (x, y - 1) are fertilized while (x + 1, y), (x, y + 1) are not, then I can get four apples from (x, y). Now, I am wondering how many apples I can get at most in the whole orchard?
 
InputThe input contains multiple test cases. The number of test cases T (T<=100) occurs in the first line of input.
For each test case, two integers N, M (1<=N, M<=100) are given in a line, which denote the size of the map. 
OutputFor each test case, you should output the maximum number of apples I can obtain. 
Sample Input

22 23 3

 
Sample Output

832題意：一個各自要麼能種一個蘋果數，要麼可以施肥，施肥的話，會使得它的上下左右蘋果樹都加倍，求最多的可能思路：依次施肥種樹計算
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 110;int map[maxn][maxn];int n, m;int cal(int i, int j) {    int ans = 1;    if (map[i][j+1] == 0)        ans <<= 1;    if (map[i+1][j] == 0)        ans <<= 1;    if (map[i-1][j] == 0)        ans <<= 1;    if (map[i][j-1] == 0)        ans <<= 1;    return ans;}int main() {    int t;    scanf("%d", &t);    while (t--) {        scanf("%d%d", &n, &m);        memset(map, -1, sizeof(map));        int flag = 0;        for (int i = 1; i <= m; i++) {            map[1][i] = flag;            flag = !flag;        }        for (int i = 2; i <= n; i++)            for (int j = 1; j <= m; j++)                map[i][j] = !map[i-1][j];        int ans = 0;        for (int i = 1; i <= n; i++)            for (int j = 1; j <= m; j++)                 if (map[i][j])                    ans += cal(i, j);            memset(map, -1, sizeof(map));        flag = 1;        for (int i = 1; i <= m; i++) {                map[1][i] = flag;                flag = !flag;            }        for (int i = 2; i <= n; i++)            for (int j = 1; j <= m; j++)                map[i][j] = !map[i-1][j];        int tmp = 0;        for (int i = 1; i <= n; i++)            for (int j = 1; j <= m; j++)                if (map[i][j])                    tmp += cal(i, j);        printf("%d\n", max(ans, tmp));    }    return 0;}